A metal sphere of radius 15 cm has a net charge of 3.0 $ 10#8 C. (a) What is the electric field at the sphere’s surface? (b) If V ! 0 at infinity, what is the electric potential at the sphere’s surface? (c) At what distance from the sphere’s surface has the electric potential decreased by 500 V?

E = 1.20 × N/C

V = 1800 V

x = 0.058 m

Explanation:

given data

radius = 15 cm

net charge = 3 × C

electric potential decreased = 500 V

solution

we get here electric field at the sphere’s surface that is

electric field at the sphere’s surface E  =     ............1

put here value

electric field at the sphere’s surface E  =    

E = 1.20 × N/C

and

potential on surface of sphere is

V =   ................2

V =  

V = 1800 V

and

now we get distance that is x

and we know here

ΔV = V(x) - V   ..............3

substitute here value

-500V =

-500 V =

solve it we get x

x = 0.058 m

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